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\begin{document}

\noindent
{\large Partial Orders, Semilattices, and Lattices: Definitions.}

\smallskip

\begin{df} A {\em partial order} on a set $A$ is a binary
relation $R\subseteq A\times A$ that is
\begin{enumerate}
\item[(i)] reflexive: $\forall a\in A\;(a\;R\;a)$;
\item[(ii)] antisymmetric: if $a\;R\;b$, then $b\;R\;a$; and
\item[(iii)] transitive: if $a\;R\;b\;R\;c$, then $a\;R\;c$.
\end{enumerate}
The structure $\lb A;R\rb$ is called a 
{\em partially ordered set}, a {\em p. o. set}, or simply a
{\em poset}. Usually we use symbols that look like 
$\leq, \preceq, \subseteq, \sqsubseteq, \ll$ in place of $R$
to denote the partial order.
\end{df}

\begin{df} If $\lb A; \leq\rb$ is a poset and $a, b\in A$, then 
an element $c$ is the {\em greatest lower bound} of $a$ and $b$
provided that
\begin{enumerate}
\item[(i)] $c\leq a$ and $c\leq b$ ($c$ is a lower bound); and
\item[(ii)] if $d\leq a$ and $d\leq b$, then $d\leq c$
($c$ is the greatest element satisfying $(i)$).
\end{enumerate}
We write $c = {\rm glb}(a,b)$ or $c = a\wedge b$ to denote that
$c$ is the greatest lower bound of $a$ and $b$.
The definition of a least upper bound is dual to the definition
of a greatest lower bound.
We write $c = {\rm lub}(a,b)$ or $c = a\vee b$ to denote that
$c$ is the least upper bound of $a$ and $b$.
\end{df}

\begin{df} (Version 1)
A {\em meet semilattice} is a poset $\lb A;\leq \rb$ where
every pair of elements has a greatest lower bound.
A {\em join semilattice} is a poset $\lb A;\leq \rb$ where
every pair of elements has a least upper bound.
\end{df}

\begin{df} (Version 2)
A {\em meet semilattice} is an algebra $\lb A;\wedge \rb$
satisfying the laws
\begin{enumerate}
\item[(i)] $x\wedge x = x$ (idempotence);
\item[(ii)] $x\wedge y = y\wedge x$ (commutativity); and
\item[(iii)] $x\wedge (y\wedge z) = (x\wedge y)\wedge z$ (associativity).
\end{enumerate}
If $\lb A;\wedge\rb$ is a meet semilattice, then the  
binary relation 
\[\leq:=\{(a,b)\in A\times A\;|\;a = a\wedge b\}\]
is the {\em associated partial order on $A$}.
\end{df}

We have proved that if $\lb A;\wedge\rb$ is a 
meet semilattice, and $\leq$ is the associated
partial order, then $\lb A;\leq\rb$ is a poset
where the greatest lower bound of any two 
elements $a, b\in A$ is given by $c = a\wedge b$.

A {\em join semilattice} is an algebra defined the same way with
$\vee$ in place of $\wedge$. Algebraically there is no difference
between a meet semilattice and a join semilattice: both are
algebras with a binary operation that is idempotent, commutative
and associative. However, we usually define the associated
partial
order differently for a join semilattice $\lb A;\vee\rb$:
\[\leq := \{(a,b)\in A\times A\;|\;b = a\vee b\}.\]
The reason for this is: if we start with a poset
$\lb A;\leq\rb$ which has both glb's and lub's, then
we have two semilattices structures: $\lb A;\wedge\rb$
and $\lb A;\vee\rb$. Our definition of ``associated partial
order'' for meet semilattices applied to $\lb A;\wedge\rb$
recovers the original order $\leq$. But if we apply the 
defintion of ``associated partial
order'' for \underline{meet} semilattices to \underline{$\lb
A;\vee\rb$},
then we do not recover the original order, we rcover the dual
order.

\begin{df} (Version 1)
A {\em lattice} is a poset $\lb A;\leq \rb$ where
every pair of elements has a greatest lower bound
and a least upper bound.
\end{df}

\begin{df} (Version 2)
A {\em lattice} is an algebra $\lb A;\wedge,\vee \rb$ 
satisfying the laws:
\begin{enumerate}
\item[(i)] (meet laws)
\begin{enumerate}
\item[(a)] $x\wedge x = x$ (idempotence);
\item[(b)] $x\wedge y = y\wedge x$ (commutativity); and
\item[(c)] $x\wedge (y\wedge z) = (x\wedge y)\wedge z$ (associativity).
\end{enumerate}
\item[(ii)] (join laws)
\begin{enumerate}
\item[(a)] $x\vee x = x$ (idempotence);
\item[(b)] $x\vee y = y\vee x$ (commutativity); and
\item[(c)] $x\vee (y\vee z) = (x\vee y)\vee z$ (associativity).
\end{enumerate}
\item[(iii)] (absorption laws)
\begin{enumerate}
\item[(a)] $x\wedge (x\vee y) = x$; and 
\item[(b)] $x\vee (x\wedge y) = x$.
\end{enumerate}
\end{enumerate}
\end{df}

\smallskip

\noindent
Exercise. Show that the two definitions of ``lattice''
are equivalent.

\smallskip

\begin{df}
A lattice is {\em distributive} if it satisfies 
the law
\[x\wedge (y\vee z) = (x\wedge y)\vee (x\wedge z).\]
A lattice is {\em modular} if it satisfies the law
\[x\wedge (y\vee (x\wedge z)) = (x\wedge y)\vee (x\wedge z).\]
\end{df}

\noindent
Facts about lattices.
\begin{itemize}
\item A distributive lattice is modular.
\item A lattice is distributive if and only if it is isomorphic
to a lattice of the form $\lb {\cal S};\cap,\cup\rb$ where
$\cal S$ is a family of sets, $\cap$ is the operation
of set intersection, and $\cup$ is the operation
of set union.
\item The lattice of subspaces of a vector space is modular.
The lattice of normal subgroups of a group is modular.
The lattice of ideals of a ring is modular.
\item If a lattice satisfies the distributive law, then
its dual also satisfies it. Hence the distributive law
for lattices, $x\wedge (y\vee z) = (x\wedge y)\vee (x\wedge z)$,
is equivalent to its dual: 
$x\vee (y\wedge z) = (x\vee y)\wedge (x\vee z)$.
Similar remarks are true for the modular law.
\end{itemize}
\end{document}

